So let us look at an example where we apply the principle of work and energy.
So what I've drawn here is a spring that's attached to the ground and on top of the spring, this little
bucket and in the bucket is a marble say.
And when it's unstitched this spring and the length of 100 millimetres, then we push down the spring
until it has a compressed length of 50 millimeters, and then we suddenly leave it and it shoots up
and the spring pushes the marble out of the bucket and the marble leaves the bucket at a certain velocity.
And we want to know what is that velocity with which the marble leaves the bucket.
So this is an ideal problem to be solved with a principle of work and energy because the spring applies
work to the model and gravity also applies.
Work to the marble and we can then plug that work done on the marble into our principle of work and
energy equation and then we can get the final velocity.
So we know the mass of the marble is 20 grams.
We know the spring constant of the spring is three kilometres per metre.
And of course we know the acceleration of gravity is nine eight one meters per second squared.
So let us write down our equation for the principle of work and energy, that is one half of one squared,
plus some of the work done is one of M.V. two squared.
Therefore, one half M.V., one squid, and I've written the work done on this particle in read the
work done by the spring minus of K X two squared minus X one squared minus.
Now we have the work done by the gravity M.G. Y two minus one, and that is equal to our final kinetic
energy.
Now let's plug in our values and let's note that we use the right size for the work done.
One of the masses, zero point zero two kilograms, the initial velocity is zero before we leave the
spring to shoot the mobile out of the bucket.
It's at a zero velocity, so that is zero.
And note that I take the mass and KG.
I want to do everything in AICI units so that the units are consistent.
Now we have the work done by the spring nets minus one half K at like three thousand Newton Burm Kenesaw
units.
And now I want to.
Take the distance over which this spring force acts, OK, now the spring moves upward and the marble
also moves upward.
So the spring as a positive work that it does on the marble.
So because in our standard form, we have a minus before the work equation for a spring, this value
in the bracket needs to be negative so that we get a positive work.
And so therefore we have zero zero five squared, minus zero point one squared.
So it's the initial position that's 15 mm, minus 100 millimetres.
So that's the distance over which the spring does the work.
Minus an hour at the work done by the gravity zero point zero two.
That's the mass multiplied by G that's nine eight one multiplied by now in the standard form we have
a minus in front of the work equation for gravity and gravity works downwards, our marble move upwards,
marble moves upwards.
So the work done by the gravity is negative.
And so because we already have a negative in front of the equation, the value in the bracket has to
be positive so that all work done stays negative.
So what is the distance over which gravity does the work?
It does it over a distance of zero comma one miner zero zero five.
That is the final position with marble leaves the bucket minus the initial position where the spring
started to push it up, and that is equal to our second kinetic energy.
So that is one half multiplied by zero point zero two, multiplied by V two squared and V two.
That is our velocity with which the marble leaves the bucket.
That is what we are interested in.
So calculating the values solving for V two squared receive two squared is one thousand one hundred
and twenty four cm zero one nine meters squared per second squared.
And if you take the square root of that, we get a velocity of thirty three point five meters per second.
And I didn't mention this in the beginning, but I hope you thought of some of the assumptions that
we've made in this problem, because there are some assumptions that we make in order for this to work
out.
The first assumption we made is that the spring is massless and that the bucket in which the model is
put is also massless because in effect, the spring also does work on the bucket because it moves the
bucket upwards.
But we've ignored that for simplicity's sake.
Then we also ignored air resistance on the back it because the resistance of air will also remove some
energy from the system.
It would also do negative work on the bucket as the bucket moves upwards.
Right.
So we ignored the force of air resistance.
I'm going to leave it to you.
Think about what other assumptions and simplifications we've made in the system.